Question

Solve the following pair of simultaneous equations:

$\frac{x}{3}+\frac{x+y}{6}=3;\frac{y}{3}-\frac{x-y}{2}=6$

• A. $x=2,y=-5$
• B. $x=7,y=4$
• C. $x=6,y=3$
• D. $x=3,y=9$

Answer 1

The correct option is D $x=3,y=9$

$\frac{x}{3}+\frac{x+y}{6}=3\Rightarrow \frac{2x+x+y}{6}=3\Rightarrow 3x+y=18$ ...(i)
$\frac{y}{3}-\frac{x-y}{2}=6\Rightarrow \frac{2y-3x+3y}{6}=6\Rightarrow -3x+5y=36$ ...(ii)
On adding (i) and (ii), we get
$3x+y=18$
$\underset{–}{-3x+5y=36}$
$6y=54$
$\therefore y=9$
On putting $y=9$ in (i), we get
$3x+9=18\Rightarrow 3x=9\Rightarrow x=3$
$\therefore x=3,y=9$